Jack Straws
In the game of Jack Straws, a number of plastic or wooden "straws" are dumped on the table and players try to remove them one-by-one without disturbing the other straws. Here, we are only concerned with if various pairs of straws are connected by a path of touching straws. You will be given a list of the endpoints for some straws (as if they were dumped on a large piece of graph paper) and then will be asked if various pairs of straws are connected. Note that touching is connecting, but also two straws can be connected indirectly via other connected straws.
Input
Input consist multiple case,each case consists of multiple lines. The first line will be an integer n (1 < n < 13) giving the number of straws on the table. Each of the next n lines contain 4 positive integers,x1,y1,x2 and y2, giving the coordinates, (x1,y1),(x2,y2) of the endpoints of a single straw. All coordinates will be less than 100. (Note that the straws will be of varying lengths.) The first straw entered will be known as straw #1, the second as straw #2, and so on. The remaining lines of the current case(except for the final line) will each contain two positive integers, a and b, both between 1 and n, inclusive. You are to determine if straw a can be connected to straw b. When a = 0 = b, the current case is terminated.
When n=0,the input is terminated.
There will be no illegal input and there are no zero-length straws.
Output
You should generate a line of output for each line containing a pair a and b, except the final line where a = 0 = b. The line should say simply "CONNECTED", if straw a is connected to straw b, or "NOT CONNECTED", if straw a is not connected to straw b. For our purposes, a straw is considered connected to itself.
Sample Input
7
1 6 3 3
4 6 4 9
4 5 6 7
1 4 3 5
3 5 5 5
5 2 6 3
5 4 7 2
1 4
1 6
3 3
6 7
2 3
1 3
0 0 2
0 2 0 0
0 0 0 1
1 1
2 2
1 2
0 0 0
Sample Output
CONNECTED
NOT CONNECTED
CONNECTED
CONNECTED
NOT CONNECTED
CONNECTED
CONNECTED
CONNECTED
CONNECTED
题目大意：按顺序输入n个线段的两个坐标，然后多组输入判断两个线段是否是连接的（相交即为连接）。
题解：利用计算几何的知识，建立线段，如果有线段相交，就用并查集把它们连在一起，然后判断根节点是不是一个就好啦。比较简单的模板题目。
有问题欢迎━(*｀∀´*)ノ亻!指出！
#include<cstdio>
#include<algorithm>
using namespace std;
struct node{
double x,y;
};
struct d
{
node a;
node b;
}deline[];
double cross(node a,node b,node o)
{
return(a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y);
}
bool isxj(d u,d v)
{
return (cross(v.a,u.b,u.a)*cross(u.b,v.b,u.a)>=)&&
(cross(u.a,v.b,v.a)*cross(v.b,u.b,v.a)>=)&&
(max(u.a.x,u.b.x)>=min(v.a.x,v.b.x))&&
(max(v.a.x,v.b.x)>=min(u.a.x,u.b.x))&&
(max(u.a.y,u.b.y)>=min(v.a.y,v.b.y))&&
(max(v.a.y,v.b.y)>=min(u.a.y,u.b.y));
}
int father[];
int find(int x)
{
return x==father[x]?x:father[x]=find(father[x]);
}
void mix(int x,int y)
{
int xx=find(x),yy=find(y);
if(xx!=yy)
{
father[xx]=yy;
}
}
int main(){
int n;
while(scanf("%d",&n),n)
{
for(int i=;i<;i++)
{
father[i]=i;
}
for(int i=;i<=n;i++)
{
scanf("%lf %lf %lf %lf",&deline[i].a.x,&deline[i].a.y,&deline[i].b.x,&deline[i].b.y);
}
for(int i=;i<=n;i++)
{
for(int j=;j<=n;j++)
{
if(isxj(deline[i],deline[j]))
{
mix(i,j);
}
}
}
int a,b;
while(scanf("%d%d",&a,&b),a&&b)
{
if(find(a)==find(b))
printf("CONNECTED\n");
else
printf("NOT CONNECTED\n");
}
}
}